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9-1.Fluid Mechanics
hard
Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be
A
$10 cm$ per sec
B
$2.5 cm$ per sec
C
$5 \times {(4)^{1/3}}cm$ per sec
D
$5 \times \sqrt 2 \,cm$ per sec
Solution
(c)If two drops of same radius r coalesce then radius of new drop is given by R
$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$==> ${R^3} = 2{r^3} \Rightarrow R = {2^{1/3}}r$
If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto {r^2}$
$\frac{{{v_2}}}{{{v_1}}} = {\left( {\frac{R}{r}} \right)^2} = {\left( {\frac{{{2^{1/3}}r}}{r}} \right)^2}$
==> ${v_2} = {2^{2/3}} \times {v_1} = {2^{2/3}} \times (5) = 5 \times {(4)^{1/3}}m/s$
Standard 11
Physics
